I'll modify my previous answer, then will apply it to derive the Post algebra as presented in @Emil's comment above.
Let $\ T\ $ be an arbitrary finite $n$-element set ($n\ge 2$). We may label it so that it becomes $\ T=\{0\ldots n\!-\!1\}$, and let $\ \Lambda:=n-1$. Now let $\ \wedge\ $ and $\ \vee\ $ be short for $ \min \ $ and $\ \max\ $ respectively, and $\ \bigwedge\ $ and $\ \bigvee\ $ be their finite iterations. Also let $\ \lambda_a : T\rightarrow T\ $ be defined by
$$ \lambda_a(x)\ :=\ \Lambda\quad\Leftrightarrow\quad a=x$$
$$ \lambda_a(x)\ :=\ 0\quad\Leftrightarrow\quad a\ne x$$
for every $\ x\in T$. Consider arbitrary $D$-argument operation $\ f:T^D\rightarrow T\ $, where $\ D\ $ is a finite set. Then
$$ f\ =\ \bigvee_{\tau\in T^D}\ (\,f(\tau)\ \wedge\ \bigwedge_{d\in D}(\lambda_{\tau(d)}\circ\pi_d)\,)$$
where $\ \pi_d:T^D\rightarrow T\ $ is the cartesian projection for every $\ d\in D$.
Thus we have a (modified) complete set of operations--it consists of $\ n\ $ constants, $\ n\ $ comparisons, and $\ \wedge\ $, and $\ \vee$.
Now let's show that the Post set $\ \{\vee\ \ \neg\ \}\ $ is complete.
Completeness of $\ \{\vee\ \ \neg\ \}$
(The equalities below are theorems, not definitions).
First of all we get constant $\ \Lambda$; it is the maximum of the consecutive compositions of the negation:
$$\Lambda\ =\ \bigvee_{k=0}^{n-1} \bigcirc^k\neg$$
Of course we identify the constants and the constant operations. Now we get every other constant $\ c\in T$:
$$ c\ = \left(\bigcirc^{c+1} \neg\right)\circ\Lambda$$
for every $\ c\in T$.
Now it's time to obtain the comparisons. Let's start with:
$$ \lambda_0\ \ =\ \ \neg\ \circ\ \bigvee_{k=0}^{n-2} \bigcirc^k\neg$$
Next:
$$ \forall_{a\in T}\quad \lambda_a\ =\ \lambda_0\,\circ\,\bigcirc^{n-a}\,\neg$$
To complete the Post's completeness only $\ \wedge\ $ is left. But first let's define an order negation $\ \sim\,:T\rightarrow T\ $ to the Post's algebraic negation $\ \neg\,$:
$$ \sim\ \ =\ \ \bigvee_{a=0}^{n-1}\ \left(\left(\bigcirc^{n-a}\ \neg\right)\,\circ\,\lambda_a\right) $$
Now the last touch is provided by the De Morgan law:
$$x\,\wedge\,y\ =\ \sim\left(\,\sim\!(x)\ \vee\,\sim\!(y)\,\right)$$
REMARK One could use exponent $\,\ -\!a\ \,$ instead of $\,\ n-a$.